5 ms·

Thanks for the numbers. 30% of delta-V but much more than 30% more fuel used (etc) since "compound interest" applies as you're using fuel to ship fuel out of o

Thanks for the numbers. 30% of delta-V but much more than 30% more fuel used (etc) since "compound interest" applies as you're using fuel to ship fuel out of orbit and then much of that fuel again to change your orbit to mars, etc. Similarly, Napoleon came up against the limits of using horses to pack hay for horses invading Russia - compound interest; you can only do so much. PS - that 30% assumes a path that I don't, and no solar power used to catapult oxygen off the moon. I'd go for a Lagrange point you're ignoring. PPS - you've assumed Mars, not, say, near-earth asteroids or trojans with a very different, far lower delta-V. Vastly different problem.

logfromblammo 8y ago
Going directly from low orbit to a transfer orbit to the second body will usually save some delta-V over going to escape orbit first and then going to transfer orbit. I didn't have the numbers for low lunar orbit to Mars transfer orbit, so I changed the Earth to Mars path for a fairer comparison. I didn't forget the Lagrange libration points, but decided not to include them because they were not relevant to the parent comment, and they would still require some mechanism to get propellant mass up to them. They're currently almost empty, and won't be very useful until the Moon itself has launch capacity to get water up there. Again, I was mainly responding to the parent post. The rocket equation is indeed a bear. As delta-V scales linearly, fuel requirements scale up exponentially. One way to manage it is to use stages. Another is to refuel and assemble vehicles higher up in the gravity well. As it happens, shipping tanks of fuel and propellant up to LEO separately means that missions to Mars starting fully fueled from LEO would require only 0.9 km/s more than missions starting from the lunar surface. Measuring by delta-V, LEO is at about the halfway point for an Earth-Mars mission. But it's also worth noting that the surface of the Moon is closer to LEO than the surface of the Earth. Shipping fuel and propellant from the Moon to LEO, to rendezvous with and refuel a vehicle coming up from Earth, is even cheaper than shipping up the resupply mass from Earth.

logfromblammo 8y ago

Going directly from low orbit to a transfer orbit to the second body will usually save some delta-V over going to escape orbit first and then going to transfer orbit. I didn't have the numbers for low lunar orbit to Mars transfer orbit, so I changed the Earth to Mars path for a fairer comparison. I didn't forget the Lagrange libration points, but decided not to include them because they were not relevant to the parent comment, and they would still require some mechanism to get propellant mass up to them. They're currently almost empty, and won't be very useful until the Moon itself has launch capacity to get water up there. Again, I was mainly responding to the parent post. The rocket equation is indeed a bear. As delta-V scales linearly, fuel requirements scale up exponentially. One way to manage it is to use stages. Another is to refuel and assemble vehicles higher up in the gravity well. As it happens, shipping tanks of fuel and propellant up to LEO separately means that missions to Mars starting fully fueled from LEO would require only 0.9 km/s more than missions starting from the lunar surface. Measuring by delta-V, LEO is at about the halfway point for an Earth-Mars mission. But it's also worth noting that the surface of the Moon is closer to LEO than the surface of the Earth. Shipping fuel and propellant from the Moon to LEO, to rendezvous with and refuel a vehicle coming up from Earth, is even cheaper than shipping up the resupply mass from Earth.